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A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.). A well-known example is the quadratic formula
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.
[1] Elementary algebra, also known as high school algebra or college algebra, [2] encompasses the basic concepts of algebra. It is often contrasted with arithmetic : arithmetic deals with specified numbers , [ 3 ] whilst algebra introduces variables (quantities without fixed values).
In mathematics, a sum of radicals is defined as a finite linear combination of n th roots: =, where , are natural numbers and , are real numbers.. A particular special case arising in computational complexity theory is the square-root sum problem, asking whether it is possible to determine the sign of a sum of square roots, with integer coefficients, in polynomial time.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
For each integer n > 2, the function n x is defined and increasing for x ≥ 1, and n 1 = 1, so that the n th super-root of x, , exists for x ≥ 1. However, if the linear approximation above is used, then y x = y + 1 {\displaystyle ^{y}x=y+1} if −1 < y ≤ 0 , so y y + 1 s {\displaystyle ^{y}{\sqrt {y+1}}_{s}} cannot exist.
The problems involve arithmetic, algebra and geometry, including mensuration. The topics covered include fractions, square roots, arithmetic and geometric progressions, solutions of simple equations, simultaneous linear equations, quadratic equations and indeterminate equations of the second degree. [10] [12]